sean is in the water 1 miles off the coast and wants to get to a town 5 miles down the coast. He needs to sim to the shore and then walk to town. If sean can swim 4 miles per hour and walk 6 miles per hour, to what point, x, should he swim to along the shoreline to reach town in least amount of time?
There are two boundary values for x:
- Sean can swim 1 mile towards the shore at a right angle to the coast, then walk 5 miles along the coast towards the town. Call the point where Sean meets the coast the x = 0 mile mark.
- Sean can swim sqrt(12 + 52) = sqrt(26) miles directly towards the town and walk 0 miles along the shore. Call the town on the coast the x = 5 mile mark.
For any value of x between 0 and 5, Sean could swim sqrt(12 + x2) = sqrt(1 + x2) miles towards the coast at some angle [spending sqrt(1 + x2)/4 hours], then walk along the coast 5 - x miles towards town [spending (5-x)/6 hours]. The time it would take Sean to travel this route would be T(x) = sqrt(1 + x2)/4 + (5 - x)/6 hours.
To minimize the time, we will find the critical numbers for the function T, and compare their function values to the boundary values.
The derivative of T(x) is
T'(x) = x / [4 sqrt(1 + x2)] - 1/6 = [3x - 2 sqrt(1 + x2)] / [12 sqrt(1 + x2)]
T'(x) is always defined on [0,5], so critical numbers for T will be x-values in [0,5] where T'(x) = 0. This will happen when
3x - 2 sqrt(1 + x2) = 0
3x = 2 sqrt(1 + x2)
9x2 =4 (1 + x2)
5x2 = 4
x = sqrt(4/5), which is about 0.894 miles.
By the Extreme Value Theorem, the minimum function value of T on [0,5] will happen either at critical numbers or at boundary values. We check their times:
T(0) = 1/4 + 5/6 = 13/12 hours,
or 1 hour and 5 minutes.
T(sqrt(4/5)) = sqrt(1 + 4/5)/4 + (5 - sqrt(4/5))/6 hours, which is about 1.02 hours,
or about 1 hour, 1 minute and 11 seconds.
T(5) = sqrt(26)/4 + 0/6 = sqrt(26)/4 hours, which is about 1.27 hours,
or about 1 hour, 16 minutes and 29 seconds.
Sean should swim directly towards the sqrt(4/5) mile mark on the coast.
